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Thread: Limit_Problem Ex_RP

  1. #1
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    Limit_Problem Ex_RP

    $\displaystyle lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n*(\frac{2n+1}{n+1})^n$
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  2. #2
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    Note that:

    $\displaystyle \left(\frac{n+1}{2n}\right) \left(\frac{2n+1}{n+1}\right)$

    $\displaystyle =\frac{2n+1}{2n}=1+\frac{1}{2n}$

    $\displaystyle \lim_{n\to \infty}\left(\frac{2n+1}{2n}\right)^{n}=\lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$

    Notice the similarity to the famous 'e' limit?.

    $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{n}\right)^{n}=e$

    So, what is the limit of $\displaystyle \lim_{n\to \infty}\left(1+\frac{1}{2n}\right)^{n}$
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  3. #3
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    Hint :
    $\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$
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  4. #4
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    Quote Originally Posted by Raoh View Post
    Hint :
    $\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n= ?$

    $\displaystyle e^{\frac{1}{2}}$

    but, what about :

    $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n$
    Last edited by gilyos; Dec 19th 2009 at 11:33 AM.
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  5. #5
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    Quote Originally Posted by Raoh View Post
    Hint :
    $\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=? $

    $\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

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  6. #6
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    Quote Originally Posted by Raoh View Post
    $\displaystyle \lim_{n\to \infty }\ln(1+\frac{1}{2n})^n=\frac{1}{2}$

    therefore your limit must be $\displaystyle e^{\frac{1}{2}}$.
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  7. #7
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    O.k but what about
    $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
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  8. #8
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    Quote Originally Posted by gilyos View Post
    O.k but what about
    $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
    i believe
    $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =0$
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  9. #9
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    $\displaystyle
    \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n = 0
    $
    Why?

    Intuition:

    $\displaystyle
    (\frac{1}{2} + \frac{1}{2n})
    $
    is always less than 1, and then raise it to any power greater than one, and it will get smaller.
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  10. #10
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    Quote Originally Posted by gilyos View Post
    O.k but what about
    $\displaystyle \lim_{n \to \infty} (\frac{1}{2}+\frac{1}{2n})^n =???$
    Given the discussion in this thread, a bit of thought might have suggested writing $\displaystyle \left( \frac{1}{2} + \frac{1}{2n}\right)^n = \left( \frac{1}{2}\right)^n \left( 1 + \frac{1}{n}\right)^n$ from which the limit is obvious.
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